Problem: Solve for $x$ : $6x^2 + 120x + 600 = 0$
Dividing both sides by $6$ gives: $ x^2 + {20}x + {100} = 0 $ The coefficient on the $x$ term is $20$ and the constant term is $100$ , so we need to find two numbers that add up to $20$ and multiply to $100$ The number $10$ used twice satisfies both conditions: $ {10} + {10} = {20} $ $ {10} \times {10} = {100} $ So $(x + {10})^2 = 0$ $x + 10 = 0$ Thus, $x = -10$ is the solution.